Hi there I am trying to accomplish something similar to the mid function in excel with a column in a pandas dataframe in python. I have a column with medication names + strengths, etc of variable length. I just want to pull out the first "part" of the name and place the result into another column in the dataframe.
Example:
Dataframe column
MEDICATION_NAME acetaminophen 325 mg a-hydrocort 100 mg/2 ml
Desired Result
MEDICATION_NAME GENERIC_NAME acetaminophen 325 mg acetaminophen a-hydrocort 100 mg/2 ml a-hydrocort
What I have tried
df['GENERIC_NAME'] = df['MEDICATION_NAME'].str[:df['MEDICATION_NAME'].apply(lambda x: x.find(' '))]
Basically I want to apply the row specific result of
df['GENERIC_NAME'] = df['MEDICATION_NAME'].apply(lambda x: x.find(' '))
to the
str[:]function?
Thanks
You can use str.partition
[ pandas-doc
] here:
df['GENERIC_NAME'] = df['MEDICATION_NAME'].str.partition(' ')[0]
For the given column this gives:
>>> g.str.partition(' ')[0]
0 acetaminophen
1 a-hydrocort
Name: 0, dtype: object
partition
itself creates from a series a dataframe with three columns: before, match, and after :
>>> df['MEDICATION_NAME'].str.partition(' ')
0 1 2
0 acetaminophen 325 mg
1 a-hydrocort 100 mg/2 ml
DO with str.split
df['MEDICATION_NAME'].str.split(n=1).str[0]
Out[345]:
0 acetaminophen
1 a-hydrocort
Name: MEDICATION_NAME, dtype: object
#df['GENERIC_NAME']=df['MEDICATION_NAME'].str.split(n=1).str[0]
Use str.extract
to use full regex features:
df["GENERIC_NAME"] = df["MEDICATION_NAME"].str.extract(r'([^\s]+)')
This capture the first word bounded by space. So will protect against instances where there are a space first.
尝试这个:
df['GENERIC_NAME'] = df['MEDICATION_NAME'].str.split(" ")[0]
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